Questions & Answers

Question

Answers

A) Line segment

B) Straight line

C) Circle

D) None of these

Answer
Verified

After separating the term ‘t’ we will take the modulus function of both the side and will eliminate ‘t’ using$ |t| = 1$.

Given data:

$

a,\;b,\;c,\; \text{and}\:{\text{ t}}\:{\text{are }}\:{\text{complex }}\:{\text{numbers }} \\

|t| = 1 \\

|t| \ne |c| \\

z = \dfrac{{at + b}}{{t - c}},z = x + iy. \\

$

Now solving the given equation, i.e.

$

z = \dfrac{{at + b}}{{t - c}} \\

\Rightarrow z(t - c) = at + b \\

\Rightarrow zt - at = b + cz \\

\Rightarrow t(z - a) = b + cz \\

\Rightarrow t = \dfrac{{b + cz}}{{z - a}} \\

$

Now, applying the modulus function on both sides of the equation, we obtain

$

\left| t \right| = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\

\sin ce{\text{ }}\left| t \right| = 1 \\

1 = \left| {\dfrac{{b + cz}}{{z - a}}} \right| \\

$

${\text{Now have also condition on}}$ ${\text{'c'}}$that \[\left| {\text{c}} \right| \ne {\text{1}}\]

$

1 = \left| {\dfrac{{c\left( {\dfrac{b}{c} + z} \right)}}{{z - a}}} \right| \\

we{\text{ }}know{\text{ }}that,{\text{ }}\left| {ab} \right| = \left| a \right|\left| b \right| \\

\Rightarrow 1 = \left| {\dfrac{{\dfrac{b}{c} + z}}{{z - a}}} \right|\left| c \right| \\

\Rightarrow \dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + \dfrac{b}{c}}}{{z - a}}} \right| \\

$

Since b, c are complex numbers, then $\dfrac{b}{c}$ be also a complex number let ‘p’

$\dfrac{1}{{\left| c \right|}} = \left| {\dfrac{{z + p}}{{z - a}}} \right|$

We know that,

$\left| {\dfrac{{z - \alpha }}{{z - \beta }}} \right| = A$

When $A \ne 1$, then the locus of z is a circle.

Here also $\left| c \right| \ne 1$ i.e.,

$\left| {\dfrac{{z + p}}{{z - a}}} \right| \ne 1$

Therefore the locus of z is a circle.

From the equation

$z = \dfrac{{at + b}}{{t - c}}$

Applying the modulus function on both sides of the equation, we obtain

$\left| z \right| = \left| {\dfrac{{at + b}}{{t - c}}} \right|$

We can say that the right-hand side will come to be a constant as a, b, c and t are some fixed complex number.

We can write it as

$\left| z \right| = f$, where f is constant, comes out to be an equation of a circle.

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